CBSE Class 10 Answered

Let P(x₁,y₁) and Q(x₂,y₂) be the points,

then using distance formula,

AP = PQ = QB

AP² = PQ² = QB²

(y₁-1)²+(x₁-2)² = (y₂-y₁)²+(x₂-x₁)² = (-8-y₂)²+(5-x₂)²  

PQ² = QB²

(y₂-y₁)²+(x₂-x₁)² = (-8-y₂)²+(5-x₂)² 

y₂² - 2y₁y₂ + y₁² + x₂² - 2x₁x₂ + x₁² =  y₂² + 16y₂ + 64 + x₂² - 10x₂ + 25

y₂² - 2y₁y₂ + y₁² + x₂² - 2x₁x₂ + x₁² =  y₂² + 16y₂ + 89 + x₂² - 10x₂

Comparing the coefficients of y₂ and x₂, on both sides,

-2y₁ = 16 and -2x₁ = -10

Hence, y₁ = -8, x₁ = 5,

P(5,-8), and it lies on the line 2x-y-k = 0,

2(5)-(-8) = k

k = 18

Regards,

Team,

TopperLearning.